ON EXISTENCE OF WEAK SOLUTIONS OF NEUMANN PROBLEM FOR QUASILINEAR ELLIPTIC EQUATIONS INVOLVING p-LAPLACIAN IN AN UNBOUNDED DOMAIN Trinh Thi Minh Hang and Hoang Quoc Toan Abstract.. In th
Trang 1ON EXISTENCE OF WEAK SOLUTIONS OF NEUMANN PROBLEM FOR QUASILINEAR ELLIPTIC EQUATIONS
INVOLVING p-LAPLACIAN IN AN UNBOUNDED DOMAIN
Trinh Thi Minh Hang and Hoang Quoc Toan
Abstract In this paper we study the existence of non-trivial weak
so-lutions of the Neumann problem for quasilinear elliptic equations in the
form
−div(h(x)|∇u| p−2 ∇u) + b(x)|u| p−2 u = f (x, u), p ≥ 2
in an unbounded domain Ω ⊂ R N , N ≥ 3, with sufficiently smooth
bounded boundary ∂Ω, where h(x) ∈ L1
loc(Ω), Ω = Ω∪ ∂Ω, h(x) ≥ 1
for all x ∈ Ω The proof of main results rely essentially on the arguments
of variational method.
1 Introduction and preliminaries results
We are concerned with the study of a Neumann problem of the type (1.1)
−div(h(x)|∇u| p −2 ∇u) + b(x)|u| p −2 u = f (x, u) in Ω,
∂u
∂n= 0 on ∂Ω, u(x) → 0 as |x| → +∞,
where p ≥ 2, Ω ⊂ R N , N ≥ 3, is an unbounded domain with sufficiently
smooth bounded boundary ∂Ω, Ω = Ω ∪ ∂Ω, n is the outward unit normal to
∂Ω, f : Ω × R −→ R is a function which will be specified later, h(x) and b(x)
are satisfied the following conditions:
(H) h(x) ∈ L1
loc(Ω), h(x) ≥ 1 for all x ∈ Ω.
(B) b(x) ∈ L ∞
loc(Ω), b(x) ≥ b0> 0 for all x ∈ Ω.
We first make some comments on the problem (1.1) In the case when Ω is
a bounded domain in RN or h(x) = 1 there were extensive studies in the last
decades dealing with the Neumann problems of type (1.1) We just remember the papers [1, 2, 4, 3], [10, 12, 13, 16], where different techniques of finding
Received June 11, 2010.
2010 Mathematics Subject Classification 35J20, 35J65.
Key words and phrases Neumann problem, p-Laplacian, Mountain pass theorem, the
weakly continuously differentiable functional.
Research supported by the National Foundation for Science and Technology Development
of Vietnam (NAFOSTED).
c
⃝2011 The Korean Mathematical Society
Trang 2solutions are illustrated We also find that in the case that h(x) ∈ L1
loc(Ω), the quasilinear elliptic equations of type (1.1), with Dirichlet boundary condition, have been studied by D M Duc, N T Vu ([7]), H Q Toan, N Q Anh, N T Chung (see [15, 14, 5]) The goal of this work we study the existence of weak solutions of Neumann problem for quasilinear elliptic equations with singular
coefficients involving the p-Laplace operator of type (1.1) in an unbounded
domain Ω⊂ R N with sufficiently smooth bounded boundary ∂Ω.
In order to state our main results let us introduce following some hypotheses:
(F1) f (x, t) ∈ C1(Ω× R, R), f(x, 0) = 0, x ∈ Ω.
(F2) There exist functions τ : Ω −→ R, τ(x) ≥ 0 for x ∈ Ω and constant
r ∈ (p − 1, N +p
N −p) such that
|f z ′ (x, z) | ≤ τ(x)|z| r −1 for a.e x ∈ Ω,
τ (x) ∈ L ∞(Ω)∩ L r0(Ω), r0= N p
N p − (r + 1)(N − p) .
(F3) There exists µ > p such that
0 < µF (x, z) = µ
∫ z
0
f (x, t)dt ≤ zf(x, z), x ∈ Ω, z ̸= 0.
Denote by
C ∞
0 (Ω) ={u ∈ C ∞ (Ω) : supp u compact ⊂ Ω}
and W 1,p(Ω) is the usual Sobolev space which can be defined as the completion
of C ∞
0 (Ω) under the norm
||u|| =
(∫
Ω (|∇u| p+|u| p )dx
)1
p
We now consider following subspace of W 1,p (Ω), defined by
H =
{
u ∈ W 1,p(Ω) :
∫ Ω
(h(x) |∇u| p + b(x) |u| p )dx < + ∞
}
and H can be endowed with the norm
||u|| H =
(∫
Ω
h(x) |∇u| p + b(x) |u| p dx
)1
p
Applying the method as those used in [14] or [5], we can prove that:
Proposition 1.1 H is a Banach space The embedding continuous H , →
W 1,p (Ω) holds true.
Proof It is clear that H is a normed space Let {u m } be a Cauchy sequence
in H Then
lim
m,k →∞
∫ Ω
(h(x) |∇(u m − u k)| p + b(x) |u m − u k | p )dx = 0
and {||u || } is bounded.
Trang 3Since||u m − u k || W 1,p(Ω)≤ b||u m − u k || H , b is a positive constant for all m, k,
{u m } is also a Cauchy sequence in W 1,p (Ω) and it converges to u in W 1,p(Ω), i.e.,
lim
∫ Ω (|∇u m − ∇u| p+|u m − u| p )dx = 0.
It follows the sequence {∇u m } converges to ∇u and {u m } converges to u in
L p(Ω) Therefore {∇u m (x) } converges to ∇u(x) and {u m (x) } converges to {u(x)} for almost everywhere x ∈ Ω Applying Fatou’s lemma we get
∫
Ω
(h(x) |∇u| p +b(x) |u| p )dx ≤ lim
m →+∞inf
∫ Ω
(h(x) |∇u m | p +b(x) |u m | p )dx < + ∞.
Hence u ∈ H Applying again Fatou’s lemma
0≤ lim
∫
Ω
(h(x) |∇u m − ∇u| p + b(x) |u m − u| p )dx
≤ lim
[
lim
k →+∞inf
∫ Ω
(h(x) |∇u m − ∇u k | p + b(x) |u m − u k | p )dx
]
= 0.
Hence{u m } converges to u in H Thus H is a Banach space and the continuous
Definition 1.1 A function u ∈ H is a weak solution of the problem (1.1) if
and only if
∫
Ω
h(x) |∇u| p −2 ∇u∇φdx +
∫ Ω
b(x) |u| p −2 uφdx −
∫ Ω
f (x, u)φdx = 0
(1.2)
for all φ ∈ C ∞
0 (Ω)
Remark 1.1 If u0∈ C ∞
0 (Ω) satisfied the condition (1.2), hence u0is a classical
solution of the problem (1.1) Indeed, since u0 ∈ C ∞
0 (Ω), supp u0 compact,
hence there exists R > 0 large enough such that ∂Ω ⊂ B R (0), supp u0 ⊂
Ω∩ B R (0) where B R (0) is ball of radius R.
By denote ΩR= Ω∩ B R(0), then from (F1) we have
∫
ΩR
h(x) |∇u0| p −2 ∇u0∇φdx +
∫
ΩR b(x) |u0| p −2 u
0φdx −
∫
ΩR
f (x, u0)φdx = 0
for all φ ∈ C ∞
0 (Ω)
Applying Green’s formula and remark that supp u0⊂ Ω ∩ B R(0) we get
∫
ΩR
−div(h(x)|∇u0| p −2 ∇u0)φ + b(x)|u0| p −2 u
0φ)dx
+
∫
∂Ω
h(x) |∇u0| p −2 ∂u0
∂n φdσ −
∫
ΩR
f (x, u0)φdx = 0 for all φ∈ C ∞
0 (Ω).
This implies that
∫
Ω
(−div (h(x)|∇u0| p −2 ∇u0)φ + b(x)|u0| p −2 u
0φ)dx −
∫ Ω
f (x, u0)φdx = 0
Trang 4for all φ ∈ C ∞
0 (ΩR ) From this it follows that
(1.3)
−div(h(x)|∇u0| p −2 ∇u0) + b(x)|u0| p −2 u0= f (x, u0) in Ω,
∂u0
∂n = 0 on ∂Ω.
Thus u0 is a classical solution of (1.1)
Our main result given by the following theorem:
Theorem 1.1 Assuming hypotheses (F1)-(F3) are fulfilled then the problem
(1.1) has at least one nontrivial weak solution in H.
Theorem 1.1 will be proved by using a variation of the Mountain pass the-orem in [6]
2 Existence of a weak solution
We define the functional J : H −→ R by
J (u) = 1
p
∫ Ω
h(x) |∇u| p dx +1
p
∫ Ω
b(x) |u| p dx −
∫ Ω
F (x, u)dx
(2.4)
= T (u) − P (u),
where
T (u) = 1
p
∫ Ω
h(x) |∇u| p dx +1
p
∫ Ω
b(x) |u| p dx
and
P (u) =
∫ Ω
F (x, u)dx.
Firstly we remark that, due to the presence of h(x) ∈ L1
loc(Ω), in general,
the functional T does not belong to C1(H) This mean that we cannot apply
the classical Mountain pass theorem by Ambrossetti-Rabinowitz In order to overcome this difficulty, we shall apply a weak version of the Mountain pass theorem introduced by D M Duc ([6]) Now we first recall the following useful concept:
Definition 2.1 Let J be a functional from a Banach space Y intoR We say
that J is weakly continuously differentiable on Y if and only if three following
conditions are satisfied:
(i) J is continuous on Y
(ii) For any u ∈ Y there exists a linear map DJ(u) from Y into R such
that
lim
t →0
J (u + tφ) − J(u)
⟨
DJ (u), φ⟩
, ∀φ ∈ Y.
(iii) For any φ ∈ Y , the map u 7→⟨DJ (u), φ⟩
is continuous on Y
Proposition 2.1 Assuming hypotheses of Theorem 1.1 are fulfilled We assert
that
Trang 5(i) P is continuous on H Moreover, P is weakly continuously differentiable
on H and
⟨
DP (u), v⟩
=
∫ Ω
f (x, u)vdx, ∀u, v ∈ H.
(ii) T is continuous on H.
(iii) T is weakly continuously differentiable on H and
⟨
DT (u), v⟩
=
∫ Ω
(
h(x) |∇u| p −2 ∇u∇v + b(x)|u| p −2 uv)
dx, ∀u, v ∈ H Thus J = T − P is weakly continuously differentiable on H and
(2.5) ⟨
DJ (u), v⟩
=
∫ Ω
(
h(x) |∇u| p −2 ∇u∇v + b(x)|u| p −2 uv)
dx −
∫ Ω
f (x, u)vdx
∀u, v ∈ H.
Proof (i) By hypotheses of Theorem 1.1, applying Theorem C1 in [11, p 248],
we have P ∈ C1(W 1,p (Ω)) Since the embedding H , → W 1,p(Ω) is continuous,
we also have P ∈ C1(H) and then P is weakly continuously differentiable on
H Moreover,
⟨
DP (u), v⟩
=
∫ Ω
f (x, u)vdx ∀u, v ∈ H.
(ii) Let{u m } be a sequence converging to u in H, i.e.,
lim
∫ Ω
(h(x) |∇u m − ∇u| p + b(x) |u m − u| p ) dx = 0.
Then{||u m || H } is bounded.
First we observe that: for some θ ∈ (0, 1):
||∇u m | p − |∇u| p | = p|∇u m + θ( ∇u m − ∇u)| p −1 |∇u m − ∇u|
≤ p2 p −2(
|∇u m | p −1 |∇u m − ∇u| + |∇u m − ∇u| p)
.
Hence by applying the Holder’s inequality we get
1
p
∫
Ω
h(x) |∇u m | p dx −1
p h(x) |∇u| p dx
(2.6)
≤ 1
p
∫
Ω
h(x) ||∇u m | p − |∇u| p |dx
≤ 2 p −2∫
Ω
h(x) |∇u m | p −1 |∇u m − ∇u|dx + 2 p −2∫
Ω
h(x) |∇u m − ∇u| p dx
≤ 2 p −2(∫
Ω
(h(x) p−1 p |∇u m | p −1) p
−1 dx
)p−1
Ω
(h(x) |∇(u m − u)| p )dx
)1
p
+ 2p −2∫
Ω
(h(x) |∇(u m − u)| p )dx
≤ c1
(
||u m || p −1
H ||u m − u|| H+||u m − u|| p
H
)
.
Trang 6Similarly, we also have
1p∫Ωb(x) |u m | p dx −1
p
∫ Ω
b(x) |u| p dx
(2.7)
≤ c2
(
||u m || p −1
H ||u m − u|| H+||u m − u|| p
H
)
.
Combining (2.6) and (2.7) we have
|T (u m)− T (u)| ≤ c3
(
||u m || p −1
H ||u m − u|| H+||u m − u|| p
H
)
with c1 , c2, c3 > 0 Letting m → +∞ since ||u m − u|| H → 0 and {||u m || H }
bounded, we obtain
lim
m →+∞ T (u m ) = T (u).
Thus T is continuous on H.
(iii) For all u, v ∈ H, any t ∈ (−1, 1) \ {0} and a.e x ∈ Ω we have
h(x) |∇u + t∇v| p − h(x)|∇u| p
t
= p
∫01h(x) |∇u + st∇v| p −2(∇u + st∇v)∇vds
≤ p
∫ 1
0
h(x) |∇u + st∇v| p −1 |∇v|ds ≤ p2 p −2 h(x)( |∇u| p −1 |∇v| + |∇v| p)
≤ p2 p −2(
h(x) p−1 p |∇u| p −1 h(x)1
p |∇v| + h(x)|∇v| p)
.
Since u, v ∈ H, we observe that
∫
Ω
(
h(x) p−1 p |∇u| p −1 h(x)1
p |∇v| + h(x)|∇v| p)
dx
≤
(∫
Ω
(h(x) p−1 p |∇u| p −1) p
−1 dx
)p−1
Ω
h(x) |∇v| p dx
)1
p
+ c5||v|| p
H
≤ c4||u|| p −1
H ||v|| H + c5||v|| p
H < + ∞,
where c4, c5 two positive constants
Hence G(x) = h(x) |∇u| p −1 |∇v|+h(x)|∇v| p ∈ L1(Ω) Applying the Lebesgue
dominated convergence theorem we get
lim
t →0
∫
Ω
h(x) |∇u + t∇v| p − h(x)|∇u| p
∫ Ω
h(x) |∇u| p −2 ∇u∇vdx.
Similarly we also have
lim
t →0
∫
Ω
b(x) |u + tv| p − b(x)|u| p
∫ Ω
b(x) |u| p −2 uvdx.
This implies that
⟨
DT (u), v⟩
= lim
t →0
T (u + tv) − T (u)
∫
(h(x) |∇u| p −2 ∇u∇v+b(x)|u| p −2 uv)dx.
Trang 7Thus T is weakly differentiable on H.
Let v ∈ H be fixed, we now prove that the map u 7→⟨DT (u), v⟩
is continuous
on H.
Assume u m → u in H, that is
lim
∫ Ω
(h(x) |∇u m − ∇u| p + b(x) |u m − u| p )dx = 0.
By hypotheses (H) and (B) it follows that∇u m → ∇u and u m → u in L p (Ω) Applying Theorem C.2 in [11, p 249] for function g(x, s) = |s| p −2 s, we deduce
that
g(x, ∇u m) =|∇u m | p −2 ∇u m −→ |∇u| p −2 ∇u
and
g(x, u m) =|u m | p −2 u
m −→ |u| p −2 u
in (L −1 p (Ω))N as m → +∞, where (L r(Ω))N = L r(Ω)× L r(Ω)× · · · × L r(Ω)
(N times) Using this fact we shall proved that the map u →⟨DT (u), v⟩
is
continuous on H for every v fixed in H.
Indeed for φ ∈ C ∞
0 (Ω), ω = suppφ, we have
|⟨DT (u m)− DT (u), φ⟩|
=
∫Ω{h(x)(|∇u m | p −2 ∇u m −|∇u| p −2 ∇u)∇φ+b(x)(|u m | p −2 u
m −|u| p −2 u)φ }dx
=
∫
ω
{h(x)(|∇u m | p −2 ∇u m −|∇u| p −2 ∇u)∇φ+b(x)(|u m | p −2 u
m −|u| p −2 u)φ }dx
≤ C(φ){||g(x, ∇u m)− g(x, ∇u)|| L p
−1 (ω) ||∇φ|| L p (ω)
+||g(x, u m)− g(x, u)|| L p
−1 (ω) ||φ|| L p (ω) },
where C(φ) is a constant positive From this letting m → +∞ we get
lim
m →+∞ |⟨DT (u m)− DT (u), φ⟩| = 0.
Since C ∞
0 (Ω) is dense in H we deduce that for every v ∈ H fixed
lim
m →+∞ |⟨DT (u m)− DT (u), v⟩| = 0.
Proposition 2.2 Suppose that sequence {u m } is weakly converging to u in
W 1,p (Ω) Then we have
T (u) ≤ lim
m →+∞ inf T (u m ).
Proof Since {u m } weakly converging in W 1,p(Ω) hence for all bounded Ω′ ⊂⊂
Ω, {u m } is also weakly converging in W 1,p(Ω′) By compactness of the em-bedding W 1,p(Ω′ ) into L p(Ω′), the sequence{u } converges strongly in L p(Ω′)
Trang 8then {u m } converges strongly in L1(Ω′) Applying Theorem 1.6 in [6, p 9] or
Theorem 4.5 [8, p 129], we deduce that
T (u) ≤ lim
m →+∞ inf T (u m ).
Proposition 2.3 The functional J : H −→ R is defined by (2.4), i.e.,
J (u) = T (u) − P (u), u ∈ H satisfies the Palais-Smale condition on H.
Proof Let {u m } be a sequence in H such that
lim
m →∞ J (u m ) = c, mlim→+∞ ||DJ(u m)|| H* = 0.
First, we shall proved that{u m } is bounded in H We suppose by contradiction
that{u m } is not bounded in H Then there exists a subsequence {u m k } of {u m }
such that ||u m k || H → +∞ as k → +∞ Observe further that
J (u m k)− 1
µ
⟨
DJ (u m k ), u m k
⟩
= T (u m k)− 1
µ
⟨
DT (u m k ), u m k
⟩ +1
µ
⟨
DP (u m k ), u m k
⟩
−P (u m k)
≥ (1
p − 1
µ)||u m k || p
H
yields
J (u m k)≥ (1
p −1
µ)||u m k || p
H+1
µ
⟨
DJ (u m k ), u m k
⟩
≥ (1
p −1
µ)||u m k || p
H −1
µ ||DJ(u u mk)|| H*||u m k || H
≥ ||u m k || H
(
γ0||u m k || p −1
µ ||DJ(u m k)|| H*
)
,
where γ0= 1p − 1
µ > 0.
From this letting k → +∞, since ||u m k || H → +∞, ||DJ(u m k)|| H* → 0, we
deduce J (u m k)→ +∞ yields a contradiction Hence {u m } is bounded in H.
By the continuous embedding H into W 1,p(Ω),{u m } is bounded in W 1,p (Ω).
Therefore, there exists a subsequence {u m k } of {u m } converging weakly to
u in W 1,p (Ω) Since the embedding W 1,p (Ω) , → L p*
(Ω) is continuous, the subsequence {u m k } converges weakly to u in L p*
(Ω) and u m k → u a.e x ∈ Ω.
It follows that {u m k } is bounded in L p*
(Ω), that is there exists a constant
M > 0 such that
||u m || ≤ M for all k = 1, 2,
Trang 9We remark that by hypotheses (F2) and (F3) we get
0≤ F (x, z) ≤ τ(x)|z| r+1 for x ∈ Ω, z ∈ R − {0},
where τ (x) ∈ L r0(Ω)∩ L ∞(Ω).
Then by Holder’s inequality and remark that 1
r0 +r+1
p* = 1 we deduce
P (u m k) =
∫ Ω
F (x, u m k )dx ≤
∫ Ω
τ (x) |u m k | r+1
≤ ||τ(x)|| L r0(Ω)||u m k || r+1
L p*(Ω)
≤ M r+1 ||τ(x)|| L r0(Ω).
By Proposition 2.2 we get
T (u) ≤ lim
k →+∞ inf T (u m k)≤ lim
k →+∞ [P (u m k ) + J (u m k)]
≤ c + ||τ(x)|| L r0(Ω)M r+1 < + ∞.
Thus u ∈ H.
Since {u m k } is weakly converges to u in L p*(Ω) and u m k → u a.e x ∈ Ω.
Then it is clear that |u m k | r −1 u
m k is converges weakly to |u| r −1 u in L p*
r (Ω)
With similar arguments as those in [9], we define the map K(u) : L p* r (Ω)−→ R
by
⟨
K(u), ω⟩
=
∫ Ω
τ (x)uωdx for ω ∈ L p*
r (Ω).
We remark that K(u) is linear and continuous provided that τ (x) ∈ L r0(Ω),
u ∈ L p*(Ω), ω ∈ L p*
r (Ω) and r1
0+p1* +p r* = 1 Hence
⟨
K(u), |u m k | r −1 u
m k
⟩
−→⟨K(u), |u| r −1 u⟩
as k → +∞,
i.e.,
k →+∞
∫ Ω
τ (x) |u m k | r −1 u
m k udx =
∫ Ω
τ (x) |u| r+1 dx.
Similarly we also have
∫ Ω
τ (x) |u m k | r+1 dx =
∫ Ω
τ (x) |u| r+1 dx.
Combining (2.8), (2.9) we get
∫ Ω
τ (x) |u m k | r −1 u
m k (u m k − u)dx = 0.
By (2.10), (F1), (F2) we obtain
lim
∫ Ω
f (x, u m k )(u m k − u)dx = 0,
i.e.,
k →+∞
⟨
DP (u m k ), u m k − u⟩= 0.
Trang 10It follows from (2.11) that
lim
⟨
DT (u m k ), u m k − u⟩= lim
⟨
DJ (u m k ), (u m k − u)⟩
+ lim
⟨
DP (u m k ), (u m k − u)⟩= 0 Moreover, since T is convex we have
T (u) − T (u m k)≥⟨DT (u m k , u − u m k)⟩
.
Letting k → +∞ we obtain that
T (u) − lim
k →+∞ T (u m k) = lim
k →+∞ [T (u) − T (u m k)]
≥ lim
⟨
DT (u m k ), u − u m k
⟩
= 0.
Thus
T (u) ≥ lim
k →+∞ T (u m k ).
On other hand, by Proposition 2.2 we have
T (u) ≤ lim
k →+∞ inf T (u m k ).
Hence, from two above inequalities, we get T (u) = lim k →+∞ T (u m k ).
Now, we shall prove that the subsequence {u m k } converges strongly to u in
H, i.e., lim k →+∞ ||u m k − u|| H = 0.
Indeed, we suppose by contradiction that{u m k } does not converge strongly
to u in H Then there exist a constant ε0 > 0 and a subsequence {u m kj } of {u m k } such that ||u m kj − u|| H ≥ ε0 for any j = 1, 2,
By recalling the Clarkson’s inequality
| α + β
2 | p+| α − β
2 | p ≤1
2(|α| p+|β| p ), ∀α, β ∈ R.
We deduce that
1
2T (u) +
1
2T (v) − T ( u + v
2 )≥ T ( u − v
2 ), ∀u, v ∈ H.
From this, for any j = 1, 2, , we have
1
2T (u m kj) +1
2T (u) − T ( u m kj + u
2 )≥ T ( u m kj − u
Remark that
T ( u m kj − u
1
p2 p ||u m kj − u|| p
p2 p ε p0.
We get
2T (u m kj) +1
2T (u) − T ( u m kj + u
p2 p ε p0.
... 0.Indeed, we suppose by contradiction that{u m k } does not converge strongly
to u in H Then there exist a constant ε0 > and a subsequence...
(N times) Using this fact we shall proved that the map u →⟨DT (u), v⟩
is
continuous on H for every v fixed in H.
Indeed for φ ∈ C ∞... sequence{u } converges strongly in L p(Ω′)
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